A292443 a(n) = (5/32)*A000045(6*n)^2.

0, 10, 3240, 1043290, 335936160, 108170400250, 34830532944360, 11215323437683690, 3611299316401203840, 1162827164557749952810, 374426735688279083601000, 120564246064461307169569210, 38821312806020852629517684640, 12500342159292650085397524884890

Offset

0,2

Comments

Every term is a triangular number. [Problem B-967 in Euler and Sadek, 2003; solution in Euler and Sadek, 2004]

Links

G. C. Greubel, Table of n, a(n) for n = 0..395

Russ Euler and Jawad Sadek, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 41, No. 5 (2003), pp. 466-471.

Russ Euler and Jawad Sadek, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 42, No. 3 (2004), pp. 277-282.

Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Formula

From Colin Barker, Sep 16 2017: (Start)

G.f.: 10*x*(1 + x) / ((1 - x)*(1 - 322*x + x^2)).

a(n) = ((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)^2) / 32.

a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2.

(End)

a(n) = A000217(A132584(n)). - Amiram Eldar, Jan 11 2022

a(n) = 10*A298101(n). - Pontus von Brömssen, Jan 06 2025

Mathematica

Table[(5/32) Fibonacci[6 n]^2, {n, 0, 13}] (* Michael De Vlieger, Sep 16 2017 *)
LinearRecurrence[{323, -323, 1}, {0, 10, 3240}, 20] (* Harvey P. Dale, Aug 31 2024 *)

Prog

(PARI) a(n) = (5/32)*fibonacci(6*n)^2
(Magma) [5*Fibonacci(6*n)^2/32: n in [0..20]]; // G. C. Greubel, Feb 03 2019
(Sage) [5*fibonacci(6*n)^2/32 for n in (0..20)] # G. C. Greubel, Feb 03 2019
(GAP) List([0..20], n-> 5*Fibonacci(6*n)^2/32); # G. C. Greubel, Feb 03 2019

Crossrefs

Cf. A000045, A132584, A298101.

Subsequence of A000217.

Sequence in context: A225764 A243008 A133198 * A001329 A007101 A007103

Adjacent sequences: A292440 A292441 A292442 * A292444 A292445 A292446

Keyword

nonn,easy

Author

Felix Fröhlich, Sep 16 2017

Status

approved